Theorem 8.3.3: Weierstrass Convergence Theorem

where || fn ||D is the sup-norm on
D. Then the (function) series
fn(x)
converges absolutely and uniformly on D to a function
f. If, in addition, each fn is continuous,
the limit function f is also continuous on D.

Context

The proof will involve two steps:

we will show that the series converges pointwise to a limit function
F

the convergence to F is actually uniform

Let's get started with the first step. Fix
x D. Then the
series f(x)
is a numeric series. In fact, since
|f(x)| || f ||D that
numeric series must converge according to the
Comparison Test
to a limit L. That limit depends on x so that we
can a function

F(x) = f(x)

which is well-defined (pointwise) for all
x D.

Next we need to show that the convergence is uniform. Take any
> 0. Since the numeric
series
|| fn ||D
converges we can find an integer N such that

<
for all n > N

But then:

which means that the sequence of partial sums converges uniformly.

If we knew what uniformly Cauchy meant, we could also show that
the sequence of partial sums was uniformly Cauchy, which (we would hope)
should imply uniform convergence. To practice, define uniformly Cauchy
sequences and finish the proof that way.